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1 month ago

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A kettle heats 1.75 kg of water. The specific latent heat of vaporisation of water is 3.34 x 106 J/kg. How much energy would be

needed to boil off the water? Write the answer out in full (i.e. not in standard form).

1 answer:

Softa [3K]1 month ago

7 0

The total energy required to evaporate the water is 5,845,000 J

Explanation:

The quantity of thermal energy needed to entirely vaporize a specific amount of a liquid substance that is already at its boiling point is represented by

$Q=m\lambda_v$

where

m indicates the mass of the substance

$\lambda_v$ represents the specific latent heat of vaporization for that substance

In this scenario, the substance in question is water, with a mass of

m = 1.75 kg

The specific latent heat of vaporization for water is

$\lambda_v = 3.34\cdot 10^6 J/kg$

Assuming the water has reached its boiling point, the thermal energy necessary to vaporize it is

$Q=(1.75)(3.34\cdot 10^6)=5,845,000 J$

Discover more about specific heat:

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Answer:

1)

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2)

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Projectile Motion

When an object is projected near the surface of the Earth at an angle $\theta$ to the horizontal, it follows a trajectory known as a parabola. The only force acting on it (ignoring wind resistance) is gravity, affecting the vertical axis.

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The vertical speed component is expressed as

$v_y=v_{oy}-gt$

1) To proceed, we will determine the initial vertical velocity component since we lack sufficient data to calculate the absolute value of $v_o$ .

The peak height is attained when $v_y=0$ , which allows us to compute the time to reach that height.

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1 month ago